Monday, March 3, 2014

I/D #2: Unit O - How can we derive the patterns for our special rights triangle?

Inquiry Activity Summary 

1. 30-60-90 Triangle

First of all, make sure you recognize that this is an equilateral triangle, whose angles sum is 180*
We cut straight down the triangle because if 180* is the total sum of the triangle's angle we are left with 30* at the top, 60* at the side, and 90* at the bottom of the triangle from 30*.
And since it is an equilateral triangle, all of the sides are of equal length of 1. But since we cut the triangle in half, the bottom side's length is cut to 1/2. With this information, we're able to find the height since the hypotenuse (c) equals 1 is already given to us and we know that one of the values is 1/2, let's call 1/2 (a) for now. If we plug these values into the Pythagorean Theorem of a^2 + b^2 = c^2, **refer to the first picture for clarification** we realize our height is radical 3/ 2. 
If we give 1 a variable instead, like n, we are left with 30* - 1/2n, 60* -  radical 3/ 2, and 90* - 1n. But we use the variable 2n so our general pattern doesn't have fractions. By doing this, we have the pattern we all know, love, and recognize of 30* - n, 60* - n radical 3, 90* - 2n. The ratio of n : 2n is to show the relationships between the sides (and like I mentioned, using the variable 2n only makes deriving easier since it takes those pesty fractions away); it only expands as the ratio remains; such as 2: 4, 3 : 6, 4: 8, etc.

2. 45-45-90 Triangle

 This time we are given an equilateral square whose sides' lengths are 1. Now a squares' total angle sum is 360* meaning each corner has an angle of 90*. We cut diagonally in half because by doing so, we cut 90* corner angle into 2 which gives 45* angles - exactly what we want! We have two sides that equal 1 and we are looking for hypotenuse's value (which is that line by which we cut the square diagonally half by.)  In order to find our hypotenuse, we shall use the Pythagorean Theorem  of a^2 + b^2 = c^2. Our two equal sides of 1 shall be a and b. **look over photo for clarification** Now by inputting 1 for a and b, our c value equals radical 2.
To find our pattern, we use a variable - which represents any number. Let's use 'n' for now. By using n to represent our two equal sides of 1, our hypotenuse now equals n radical 2; the variable does allow for the relationship to stay consistent and allows us to expand it should the case ever be that the side length equal a number greater than one (this also applies to the 30-60-90 triangle.)

Inquiry Activity Reflection

Something I never noticed before about right triangles is how we are able to find these two triangles in equilateral shapes which just allows the whole derivation to be much easier; but something that I had forgotten and was brought back to my attention was the total angle sum of these equilateral shapes that allows us to create the triangles in the first place by cutting these shapes up either diagonally or straight down the half.  
Being able to derive these triangles myself aids in my learning because heaven forbid that I should ever have a brain fart on these concepts, I'll be able to fully comprehend special rights triangle and remember how to solve problems such as concept 7-8 by going back to these basics. And also being able to understand why a 45* side value is n and why a 60* side's value is n radical 3 and not because of that cute little memory trick of the 60's being radical times, which can also be considered with the 45-45-90 triangle with the 90* side value of n radical 2 because the 90's were also radical times. 

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